Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[110][110], row[110], col[110], ansrow[110], anscol[110], n, m, tans, ans;
bool flag;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
  ans = 10000000;
  for (int i = 0; i <= 500; i++) {
    row[1] = i;
    flag = 1;
    for (int j = 1; j <= m; j++)
      if (a[1][j] < i) {
        flag = 0;
        break;
      } else
        col[j] = a[1][j] - i;
    if (!flag) break;
    for (int j = 2; j <= n; j++) {
      if (a[j][1] - col[1] < 0) {
        flag = 0;
        break;
      }
      row[j] = a[j][1] - col[1];
      for (int k = 2; k <= m; k++)
        if (a[j][k] - col[k] != row[j]) {
          flag = 0;
          break;
        }
      if (!flag) break;
    }
    if (!flag) continue;
    tans = 0;
    for (int j = 1; j <= n; j++) tans += row[j];
    for (int j = 1; j <= m; j++) tans += col[j];
    if (tans < ans) {
      ans = tans;
      for (int j = 1; j <= n; j++) ansrow[j] = row[j];
      for (int j = 1; j <= m; j++) anscol[j] = col[j];
    }
  }
  if (ans == 10000000) {
    printf("-1");
    return 0;
  }
  printf("%d\n", ans);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= ansrow[i]; j++) printf("row %d\n", i);
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= anscol[i]; j++) printf("col %d\n", i);
  return 0;
}